Hdu 4630 No Pain No Game

Question: give a certain arrangement of n and 1 to n, ask q times, and each time ask [l, r] to pick two numbers at will, what is the maximum value of the maximum common approx.

Solution: record a pre array. pre [I] indicates the position at which the multiples of a certain number of I have appeared. Sort the query by right endpoint. Use a tree array to maintain the best answer from a certain position to maxn. Scanning num [I] from 1 to n, and enumerating its approximate number x for the complexity of sqrt (num [I, then we can use this x to update all the answers before pre [x] (update when update is available, and maintain a maximum value for the suffix array ). Assume that the right endpoint of the k-th query is I, because we only process I, so the maximum value from the left endpoint to maxn is the answer to the k-th query.

 

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std ;const int maxn = 55555 ;struct Ans{int l , r , ans ;} p[maxn] ;int c[maxn] , pre[maxn] ;int lowbit ( int x ) { return x & ( -x ) ; }void update ( int pos , int v ){while ( pos > 0 ){c[pos] = max ( c[pos] , v ) ;pos -= lowbit ( pos ) ;}}int query ( int pos ){int ret = 0 ;while ( pos < maxn - 10 ){ret = max ( ret , c[pos] ) ;pos += lowbit ( pos ) ;}return ret ;}bool cmp ( int i , int j ){return p[i].r < p[j].r ;}int pos[maxn] , num[maxn] ;int main (){int cas , n , i , j , m ;scanf ( "%d" , &cas ) ;while ( cas -- ){memset ( c , 0 , sizeof ( c ) ) ;memset ( pre , 0 , sizeof ( pre ) ) ;scanf ( "%d" , &n ) ;for ( i = 1 ; i <= n ; i ++ ) scanf ( "%d" , &num[i] ) ;scanf ( "%d" , &m ) ;for ( i = 1 ; i <= m ; i ++ ){scanf ( "%d%d" , &p[i].l , &p[i].r ) ;pos[i] = i ;}sort ( pos + 1 , pos + m + 1 , cmp ) ;int tot = 1 ;for ( i = 1 ; i <= n ; i ++ ){if ( tot > m ) break ;for ( j = 1 ; j * j <= num[i] ; j ++ ){if ( num[i] % j ) continue ;if ( pre[j] != 0 ) update ( pre[j] , j ) ;pre[j] = i ;if ( j * j == num[i] ) continue ;int k = num[i] / j ;if ( pre[k] != 0 ) update ( pre[k] , k ) ;pre[k] = i ;}while ( tot <= m && p[pos[tot]].r == i ){p[pos[tot]].ans = query ( p[pos[tot]].l ) ;tot ++ ;}}for ( i = 1 ; i <= m ; i ++ ){if ( p[i].l == p[i].r ){puts ( "0" ) ;continue ;}printf ( "%d\n" , p[i].ans ) ;}}}